7. TRIGONOMETRIC RATIOS, IDENTITIES AND EQUATIONS 1. INTRODUCTION 2. TRIGONOMETRIC FUNCTIONS (CIRCULAR FUNCTIONS)

Transcript

1 7. TRIGONOMETRIC RATIOS, IDENTITIES AND EQUATIONS. INTRODUCTION The equations involving trigonometric functions of unknown angles are known as Trigonometric equations e.g. cos 0,cos cos,sin + sin cos sin.. TRIGONOMETRIC FUNCTIONS (CIRCULAR FUNCTIONS) Function Domain Range sin A R [, ] cos A R [-, ] tan A R ( n + ) /,n I R (, ) cosec A R n,n I (,, ) sec A R {( n + ) /,n I} (,, ) cot A R n,n I (, ) We find, sina, cosa,seca or seca and coseca or coseca. Some Basic Formulae of Trigonometric Functions Y (a) (b) (c) sin A + cos A. sec A tan A cosec A cot A (d) sinacoseca tanacot A cosaseca A system of rectangular coordinate axes divide a plane into four quadrants. An angle lies in one and only one of these quadrants. The signs of the trigonometric ratios in the four quadrants are shown in Fig 7.. II quadrant only sin are + ve cosec } III quadrant only tan are + ve cot } I quadrant All t-ratios are + ve X X Y Figure 7. IV quadrant only cos are + ve sec }

2 7. Trigonometric Ratios, Identities and Equations MASTERJEE CONCEPTS A crude way to remember the sign is Add Sugar to Coffee. This implies the st letter of each word gives you the trigonometric functions with a +ve sign. Eg. Add- st word st quadrant st lettera All are positive to- rd word rd quadrant st letter-t tan (cot ) are positive. Ravi Vooda (JEE 009, AIR 7) Sine, cosine and tangent of some angles less than 90º: Trigonometric ratios 0º 5º 8º 0º 6º sin 0 cos tan sin /5 7º 5º 5º 60º 90º /5 cos /5 /5 0 tan / / Not defined Illustration : Prove the following identities: (i) + tan A + + tan A sin A sin A (ii) + tan tan + cot cot Sol: (i) Simply by using Pythagorean and product identities, we can solve these problems. (i) L.H.S. ( + tan A) + + tan A sec A + ( + cot A) sec A + cosec A sin A + cos A + sin + cos cos A sin A sin A.cos A

3 Mathematics 7. R.H.S. cos sin sin A sin A sin A ( sin A) Hence proved. (ii) L.H.S. + tan sec sin tan + cot cosec cos tan tan tan Now, R.H.S. cot tan tan tan From (i) and (ii), clearly, L.H.S. R.H.S. tan ( tan ).tan tan (i) (ii) Proved. Illustration : Prove the following identities: (i) sin A cos A + sec A cosec A cos A sin A sin Acos A (ii) sec A ( sin A) tan A Sol: Use algebra and appropriate identities to solve these problems. (i) sin A cos A sin A + cos A sin A + cos A + sin Acos A sin A.cos A + cos A sin A sin Acos A sin Acos A ( sin A + cos A ) sin Acos A sin Acos A sin Acos A sin Acos A sin Acos A sec A cosec A R.H.S. Proved. sin Acos A sin Acos A (ii) L.H.S. sec A ( sin A) tan sin A A sec A tan A sec A tan A tan A cos A + tan A tan A tan A + tan A + tan A tan A tan A R.H.S. Proved. Illustration : Prove the following identities: (i) + cos α cosec α+ cot α cosα (ii) + sin α sec α+ tan α sinα Sol: By rationalizing L.H.S. we will get required result. (i) L.H.S. ( + cosα) + cosα + cosα + cosα cos cos cos cos α α + α α + cosα + cosα cosα + cosecα+ cotα R.H.S. sin α sinα sinα sinα (ii) L.H.S. + sinα + sinα + sinα + sinα + sinα sin sin sin sin cos α α + α α α Proved.

4 7. Trigonometric Ratios, Identities and Equations + sin α sin α + sec α+ tanα R.H.S. cosα cosα cosα Proved. Illustration : In each of the following identities, show that: (i) cot Α+ tanb sin A sin B cot A.tanB (ii) tan A tan B (JEE ADVANCED) cotb + tana cos Acos B Sol: Apply tangent and cotangent identity. (i) L.H.S. cosa sinb cosacosb + sinasinb + cot A + tanb sina cosb sinacosb cotb + tana cosb sina cosacosb + sinasinb + sinb cosa sinbcosa sinbcosa cosa sinb cot A tanb R.H.S. sinacosb sina cosb sin A sin B sin Acos B cos Asin B (ii) L.H.S. tan A tan B cos A cos B cos Acos B ( ) ( ) sin A sin B sin A sin B sin A sin A sin B sin B + sin A sin B cos Acos B cos Acos B Proved. sin A sin B R.H.S. Proved. cos Acos B Illustration 5: Prove the following identities: cosec cot sin sin cosec+ cot (JEE ADVANCED) Sol: By rearranging terms we will get we can solve this problem. +, and then using Pythagorean identity cosec cot cosec+ cot sin We have, cosec cot sin sin cosec+ cot cosec cot cosec+ cot sin sin cosec cot cosec+ cot sin Now, L.H.S. cosec+ cot+ cosec cot + cosec cot cosec+ cot cosec cot cosec+ cot cosec ( cosec cot ) cosec R.H.S. sin cosec cot Alternative Method ( cosec cot ) R.H.S cosec sin cosec + cot cosec cot cosec Proved. sin cosec cosec + cot cot Proved.

5 Mathematics 7.5 Illustration 6: Prove that: + cot A + tana sina cosa (i) sec A cosec A sin A.cos A sina cosa (ii) + seca + tana coseca + cot A (JEE ADVANCED) Sol: Using algebra and appropriate identities, we can prove this. ( + cot A + tana )( sina cosa ) (i) L.H.S. sec A cosec A cosa sina + + sina cosa sina cosa ( sec A coseca)( sec A + sec A cosec A + cosec A) a b a b a + ab + b sinacosa + cos A + sin A sina cosa sina cosa ( sinacosa + ) sinacosa sinacosa sinacosa sec A coseca sin A sinacosa cos A ( sec A coseca cosasina ) cos A sin A sin Acos A ( sinacosa + )( seca coseca ) ( seca coseca)( + sinacosa) sin Acos A sin + cos sin Acos A R.H.S. Proved. (ii) L.H.S. sina cosa + seca + tana cosec A + cot A sinacoseca + sinacot A sina + cosaseca + cosa tana cosa ( seca + tana )( coseca + cot A ) + cosa sina + + sina cosa sina cosa sina cosa cosa sina cosa cosa sina sina cosa sina sinacosa sinacosa sina cosa sina cosa + sina cosa sinacosa sin A + cos A sinacosa sinacosa sinacosa sinacosa + sinacosa ( a+ b)( a b) a b sin + cos sinacosa R.H.S. sinacosa Proved. Illustration 7: Prove that: sin.cos + sin.cos sec cos cosec sin + sin.cos (JEE ADVANCED) Sol: Write L.H.S. in terms of cosine and sine functions.

6 7.6 Trigonometric Ratios, Identities and Equations L.H.S. + sin.cos sec cos cosec sec cos sin + sin cos + sin cos cos sin cos sin cos sin cos sin + sin cos a b ( + cos )( cos ) ( sin )( + sin ( a b)( a+ b) ) cos sin + sin cos ( cos ) sin cos ( sin + + ) cos ( sin ) sin ( cos cos sin ) + + cos + sin ( + cos )( + sin ) cos + sin cos + sin + sin cos sin + cos + sin cos ( cos + sin ) + cos + sin + cos + sin ( + cos )( + sin ) sin + cos + sin cos sin cos sin + cos sin cos sin cos sin cos R.H.S. + sin + cos + sin cos + + sin cos + sin cos Proved.. TRANSFORMATIONS. Compound, Multiple and Sub-Multiple Angles Circular functions of the algebraic sum of two angles can be expressed as circular functions of separate angles. sin (A ± B) sin A cos B ± cos A sin B; tan A tana ± tanb tana.tanb ( ± B) ; cot ( A B) cos (A ± B) cos A cos B sin A sin B cot AcotB ± cotb ± cot A Circular functions of multiples of an angle can be expressed as circular functions of the angle. sina tana sinacosa + tan A cosa cos A sin A tan A tan A + cos A sin A tana ; tana tan A cosa cos A cosa ; sina sina sin A tana tan A tana tan A

7 Mathematics 7.7 Circular functions of half of an angle can be expressed as circular functions of the complete angle. A cosa sin ; A + cosa cos ; A cosa cosa sina tan + cosa sina + cosa. Complementary and Supplementary Angles sin cos tan sin cos cos + sin tan + cot ( ) tan sin sin sin cos cos sin tan cot sin + cos cos tan sin cos tan cos tan + sin + cos + tan. Product to Sum and Sum to Product C+ D C D sinc + sind sin.cos ; C+ D C D cosc + cosd cos.cos ; C+ D C D sinc sind cos.sin C+ D D C cosc cosd sin.sin C+ D C + D Note: sinc + cosd sinc + sin D sin.cos sinc sind sin( C + D) + + ; sina.cosb { sin( A + B) + sin( A B) } tanc tand cosc cosd cosc.cosd { ( ) ( + )} ; cosa.cosb { cos( A B) + cos( A + B) } sina.sinb cos A B cos A B sin( A + B ).sin( A B) sin A sin B ; ( + ) ( ). Power Reduction ( ) cos A ( + cosa) sin A cosa cos A B.cos A B cos A sin B cosa tan A ; + cosa sina sina sin A ; cosa + cosa cos A

8 7.8 Trigonometric Ratios, Identities and Equations MASTERJEE CONCEPTS n n sin A cosa.cosa.cos.acos A...cos A if A n n sina if A n if A (n+ ) sin( A + A A ) cosa cosa...cosa ( S S + S S +...) n n 5 7 cos( A + A A ) cosa cosa...cosa ( S + S S...) n n 6 S S + S S +... tan A + A An S + S S Where, S tan A + tan A +. + tan A The sum of the tangents of the separate angles. n S tan A tan A + tan A tan A +. The sum of the tangents taken two at a time. S tan A tan A tan A + tan A tan A tan A +.. Sum of tangents three at a time, and so on. If A A.. A n A, then S n tan A, S n C tan A. S n C tan A,.. Vaibhav Gupta (JEE 009, AIR 5). TRIGONOMETRIC IDENTITY A trigonometric equation is said to be an identity if it is true for all values of the angle or angles involved. A given identity may be established by (i) Reducing either side to the other one, or (ii) Reducing each side to the same expression, or (iii) Any convenient, modification of the methods given in (i) and (ii).. Conditional Identity When the angles, A, B and C satisfy a given relation, we can establish many interesting identities connecting the trigonometric functions of these angles. To prove these identities, we require the properties of complementary and supplementary angles. For example, if A + B + C, then. sin( B + C) sina,cosb cos( C + A). cos( A + B) cosc,sinc sin( A + B). tan( C + A) tanb,cot A cot( B + C). A+ B C C A+ B cos sin,cos sin 5. C+ A B A B+ C sin cos,sin cos 6. B+ C A B C+ A tan cot,tan cot Some Important Identities: If A+ B+ C, then. tana + tanb + tanc tana tanb tanc. cotbcotc + cotccot A + cot AcotB B C C A A B. tan tan + tan tan + tan tan. A B C A B C cot + cot + cot cot cot cot

9 Mathematics sina + sinb + sinc sina sinb sinc 6. cosa + cosb + cosc cosacosbcosc 7. cos A + cos B + cos C cosacosbcosc 8. A B C 9. cosa + cosb + cosc + sin sin sin Illustration 8: Show that: A B C sina + sinb + sinc cos cos cos (i) sin( 0º ) cos( 0º ) cos( 0º ) sin( 0º ) (ii) cos cos φ sin sin φ sin( +φ) Sol: Use sum and difference formulae of sine and cosine functions. (i) L.H.S. sin( 0º + ) cos( 0º +) cos( 0º + ) sin( 0º +) sin{ ( 0º +) ( 0º + )} ( ) sin A B sinacosb cosasinb sin0º R.H.S. Proved. (ii) L.H.S. cos cos φ sin sin φ cos + φ cos( A B) cosacosb sinasinb cos + +φ sin( +φ ) R.H.S. cos sin Proved Illustration 9: Find the value of tan ( α+β ), given that cot α, α, and 5 sec β, β,. tanα+ tanβ α+β, therefore by using product and Pythagorean identities we can obtain tan α.tan β the values of tanα and tanβ. Given, cotα tanα 5 5 Also, secβ. Then tanβ sec β ± ± 9 But β, tanβ tanβ is ve in II quadrant + + Substituting tanα and tanβ in (), we get tan ( α+β ) + Sol: As we know, tan Illustration 0: Prove that: tana tana tana tana tana tana Sol: Here we can write tana as tan( A + A), and then by using tan problem. We have: A A + A tana tan( A + A) tana + tana tana tana tana tanα+ tanβ α+β we can solve this tan α tan β

10 7.0 Trigonometric Ratios, Identities and Equations tana( tana tana) tana + tana tana tana tana tana tana + tana tana tana tana tana tana tana Proved. Illustration : Prove that: cos8 cos Sol: Use + cos cos, to solve this problem. L.H.S cos8 + + ( + cos8) + + (cos ) cos ( cos ) + cos cos cos + cos ( + cos ).cos cos cos R.H.S. Proved. Illustration : If tan A m m Sol: Simply using tan (A B) We have, tan A and tan B m m m and tan B m, prove that A B tana tanb, we can prove above equation. + tana.tanb (JEE ADVANCED) Now, tan (A B) tana tanb + tana.tanb... (i) Substituting the values of tan A and tan B in (i), we get tan (A B) m m m m + m m m m m ( m )( m ) m m m m m tan( A B) tan tan A B Proved. nsinαcosα Illustration : If tanβ nsin Sol: Same as above problem tan α ; prove that nsinαcosα tanβ, we can prove given equation. nsin α L.H.S. tan tanα tanβ α β + tan α tan β tan α β n tanα (JEE ADVANCED) tanα tanβ α β, therefore by substituting + tan α tan β sinα nsinαcosα nsinαcosα Substituting tanβ nsin α in (i), we get L.H.S. cosα nsin α sinα tanα sinα nsinαcosα cos +. α cosα nsin α... (i)

11 Mathematics 7. ( ) ( ) sinα nsin α nsinαcos α cosα nsin α + nsin αcosα sinα nsin α nsinαcos α cosα nsin αcosα+ nsin αcosα sinα nsinα sin α+ cos α sinα nsinα cosα cosα sin α+ cos α n sinα n tan α R.H.S. cosα Proved. Illustration : If +φα and sin k sinφ, prove that k sinα sinα tan,tanφ + k cosα k + cosα (JEE ADVANCED) Sol: Here φα, substitute this in sin k sinφand then use compound angle formula to obtain required result. We have, +φα φα... (i) and sin k sinφ... (ii) sin k sin( α ) [Using (i)] k sinαcos cosαsin sin k sinαcos kcosαsin... (iii) Dividing both sides of (iii) by cos, we get tan k sinα kcos α.tan k sinα tan+ k cos α.tan k sinα tan ( + k cosα ) k sinα tan Proved. + k cos α Again, sin k sinφ sin( α φ ) k sinφ +φα α φ sinαcosφ cosαsinφ k sinφ... (iv) Dividing both side of (iv) by cosφ, we get sinα sinα cosαtanφ k tanφ ( k + cosα) tanφ sinα tanφ Proved. k + cos α α+β β+γ γ+α Illustration 5: Prove that: cosα+ cosβ+ cos γ+ cos( α+β+γ ) cos cos cos (JEE ADVANCED) α+β α β Sol: Use cosα+ cosβ cos cos, to solve this problem. L.H.S cosα+ cosβ+ cos γ+ cos( α+β+γ ) ( cosα+ cosβ ) + cos γ+ cos( α+β+γ) α+β α β α+β+γ+γ α+β+γ γ cos cos + cos.cos α+β α β α+β+ γ α+β cos cos + cos cos α+β α β α+β+ γ cos cos + cos α β α+β+ γ α+β+ γ α β + α+β cos cos cos

12 7. Trigonometric Ratios, Identities and Equations α+β α+γ β+γ cos cos cos α+β β+γ γ+α cos cos cos R.H.S. Proved. Illustration 6: If xcos y cos + zcos +, then show that xy + yz + zx 0. (JEE ADVANCED) Sol: Consider xcos y cos + zcos + k, obtain the value of x, y and z in terms of k, and solve L.H.S. of given equation. Let xcos y cos + zcos + k... (i) cos + cos + cos,, x k y k z k xyz xyz xyz Now, L.H.S. xy + yz + zx + + xyz + + z x y z x y cos + cos + cos xyz xyz + + [Using (i)] cos cos cos k k k k xyz cos + cos + cos k xyz cos ( + ) cos + cos k xyz cos. + cos k xyz cos cos k + xyz 0 k 0 xy + zy + zx 0 Proved. n sin n Illustration 7: Prove that: coscoscos...cos n sin Sol: Multiply and divide L.H.S. by sin and apply sin( ) sincos. Here, we observe that each angle in L.H.S. is double of the preceding angle. n L.H.S. coscoscos...cos n n ( sin.cos) cos.cos...cos ( sin.cos)( cos...cos ) sin sin n ( sin.cos ) cos8cos6...cos sin n sin n cos n sin n n ( sin8.cos8 ) cos6...cos sin( ) n n sin cos R.H.S. sin n n sin sin (JEE ADVANCED) Proved. acosφ+ b a b Illustration 8: If cos, prove that tan tan φ (JEE ADVANCED) a + bcos φ a+ b φ tan tan Sol: Substitute cos and acosφ+ b cosφ in given equation i.e. cos. φ a + bcos φ + tan + tan

13 Mathematics 7. Now, acos b cos φ+ a + bcos φ φ tan a. + b φ tan tan + φ + tan tan a+ b φ tan + [Using (i)] φ φ tan a tan b tan + + φ φ + tan a tan b tan + + Applying componendo and dividendo, we get φ φ a atan + b + btan φ φ a + atan + b btan φ φ atan btan φ tan ( a b) tan a + b a + b a b tan tan φ a+ b Proved 5. SOLUTION OF TRIGONOMETRIC EQUATION A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. 9 Eg.: if sin,,,,... Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature). These solutions can be classified as: (i) Principal solution 5. Principal Solution (ii) General solution The solutions of a trigonometric equation which lie in the interval, ) are called principal solutions. Methods for Finding Principal Value Suppose we have to find the principal value of satisfying the equation sin. Since sin is negative, will be in rd or th quadrant. We can approach the rd and the th quadrant from two directions. Following the anticlockwise direction, the numerical value of the angle will be greater than. The clockwise approach would result in the angles being numerically less than. To find the principal value, we have to take the angle which is numerically smallest. Y + X /6 /6 X B Y Figure 7.

14 7. Trigonometric Ratios, Identities and Equations For Principal Value (a) If the angle is in the st or nd quadrant, we must select the anticlockwise direction and if the angles are in the rd or th quadrant, we must select the clockwise direction. (b) Principal value is never numerically greater than. (c) Principal value always lies in the first circle (i.e. in first rotation) 5 On the above criteria, will be or. Among these two 6 6 principal value of satisfying the equation sin has the least numerical value. Hence 6 From the above discussion, the method for finding principal value can be summed up as follows: (a) First identify the quadrants in which the angle lies. is the 6 (b) Select the anticlockwise direction for the st and nd quadrants and select clockwise direction for the rd and th quadrants. (c) Find the angle in the first rotation. (d) Select the numerically least value among these two values. The angle thus found will be the principal value. (e) In case, two angles, one with a positive sign and the other with a negative sign have the same numerical value, then it is the convention to select the angle with the positive sign as the principal value. 5. General Solution The expression which gives all solutions of a trigonometric equation is called a General Solution. General Solution of Trigonometric Equations In this section we shall obtain the general solutions of trigonometric equations sin 0,cos 0,tan 0 and cot 0. General Solution of sin 0 By Graphical approach: The graph clearly shows that sin 0 at 0,,,...,,... O - - / - - / + - Figure 7. So the general solution of sin 0 is n :n I where n 0, ±, ±.. Note: Trigonometric functions are periodic functions. Therefore, solutions of trigonometric equations can be generalized with the help of periodicity of trigonometric functions. MASTERJEE CONCEPTS A trigonometric identity is satisfied by any value of an unknown angle while a trigonometric equation is satisfied by certain values of the unknown. Vaibhav Krishnan (JEE 009, AIR ) Method for Finding Principal Value (a) First note the quadrants in which the angle lies. (b) For the st and nd quadrants, consider the anticlockwise direction. For the rd and th quadrants, consider the clockwise direction.

15 Mathematics 7.5 (c) Find the angles in the st rotation. (d) Select the numerically least value among these two values. The angle found will be the principal value. Illustration 9: Principal value of tan is Sol: Solve it by using above mentioned method. tan is negative will lie in nd or th quadrant For the nd quadrant, we will choose the anticlockwise direction and for the th quadrant, we will select the clockwise direction. In the first circle, two values and are obtained. Among these two, is numerically least angle. Hence, the principal value is Y B + X X A Y Figure 7. Illustration 0: Principal value of cos Sol: Here cos is ( + ) ve hence will lie in st or th quadrant. cos is ( + ) ve will lie in the st or the th quadrant. is: For the st quadrant, we will select the anticlockwise direction and for the th quadrant, we will select the clockwise direction. As a result, in the first circle, two values and are found. X Both and have the same numerical value. In such a case, will be selected as the principal value, as it has a positive sign. Illustration : Find the general solutions of the following equations: O Y - Y Figure 7.5 B A X (i) sin 0 (ii) cos 0 (iii) tan 0 Sol: By using above mentioned method of finding general solution we can solve these equation. n (i) We have, sin 0 n where, n 0, ±, ±, ±... n Hence, the general solution of sin 0 is,n Z (ii) We know that, the general solution of the equation cos 0 is ( n + ),n Z Therefore, cos 0 ( n + ) ( n + ), where n 0, ±, ±. Which is the general solution of cos 0 (iii) We know that the general solution of the equation tan 0 is n, n Z Therefore, tan 0 tan 0 n Which is the required solution. n, where n 0, ±, ±

16 7.6 Trigonometric Ratios, Identities and Equations 6. PERIODIC FUNCTION A function f (x) is said to be periodic if there exists T > 0 such that f (x + T) f (x) for all x in the domain of definition of f (x). If T is the smallest positive real number such that f (x + T) f (x), then it is called the period of f (x). We know that, sin( n+ x) sinx,cos( n+ x) cosx, tan n+ x tanx for all n Z Therefore, sinx, cosx and tanx are periodic functions. The period of sinx and cosx is and the period of tanx is. Function Period sin( ax + b ),cos( ax + b ),sec( ax + b ),cosec( ax + b) /a tan( ax + b ),cot( ax + b) /a sin( ax + b ), cos( ax + b ), sec( ax + b ), cosec( ax + b) /a tan( ax + b ), cot( ax + b) / a (a) Trigonometric equations can be solved by different methods. The form of solutions obtained in different methods may be different. From these different forms of solutions, it is wrong to assume that the answer obtained by one method is wrong and those obtained by another method are correct. The solutions obtained by different methods may be shown to be equivalent by some supplementary transformations. To test the equivalence of two solutions obtained from two different methods, the simplest way is to put values of n..,, 0,,,,. etc. and then to find the angles in [0, ]. If all the angles in both the solutions are same, the solutions are equivalent. (b) While manipulating the trigonometric equation, avoid the danger of losing roots. Generally, some roots are lost by cancelling a common factor from the two sides of an equation. For example, suppose we have the equation tan x sin x. Here by dividing both sides by sin x, we get cos x /. (c) While equating one of the factors to zero, we must take care to see that the other factor does not become infinite. For example, if we have the equation sin x 0, which can be written as cos x tan x 0. Here we cannot put cos x 0, since for cos x 0, tan x sin x / cos x is infinite. (d) Avoid squaring: When we square both sides of an equation, some extraneous roots appear. Hence it is necessary to check all the solutions found by substituting them in the given equation and omit the solutions that do not satisfy the given equation. For example: Consider the equation, sin+ cos. (i) Squaring, we get + sin or sin 0. (ii) This gives 0, /,, / Verification shows that and / do not satisfy the equation as sin+ cos, and sin / + cos /,. The reason for this is simple. The equation (ii) is not equivalent to (i) and (ii) contains two equations: sin+ cos and sin+ cos. Therefore, we get extra solutions. Thus if squaring is a must, verify each of the solutions. Some Necessary Restriction: If the equation involves tan x, sec x, take cos x 0. If cot x or cosec x appear, take sin x 0. If log appears in the equation, then number > 0 and base of log > 0,. Also note that f ( ) is always positive. For example, sin sin, not ± sin. Verification: Students are advised to check whether all the roots obtained by them satisfy the equation and lie in the domain of the variable of the given equation.

17 Mathematics SOME TRIGONOMETRIC EQUATIONS WITH THEIR GENERAL SOLUTIONS Trigonometric equation General solution If sin 0 n If cos 0 n + / n+ / If tan 0 n If sin n + / n+ / If cos n If sin sinα n+ ( ) n αwhere α /, / If cos cosα n±α where α 0, If tan tanα n+α where α /, / If sin If cos sin α n±α cos α n±α If tan tan α n±α If sin sinα cos cosα If sin sinα tan tanα If tan tanα cos cosα n+α n+α n+α Note: Everywhere in this chapter, n is taken as an integer. Illustration : Solve: sinm+ sinn 0 α+β α β Sol: By using sinα+ sinβ sin cos, we can solve this problem. We have, sinm+ sinn 0 m+ n m n sin.cos 0 m+ n sin 0 or m n cos 0 Now, m+ n sin 0 m+ n r,r Z... (i)

18 7.8 Trigonometric Ratios, Identities and Equations And m n cos 0 m n ( p + ) m n cos cos, p Z r From (i) and (ii), we have m+ n p +, m n or p + m where, m,n Z (ii) Illustration : Solve: sinxcosx + sinx + cosx + 0 (JEE ADVANCED) Sol: Simply using algebra and method of finding general equation, we can solve above equation. We have, sinxcosx + sinx + cosx + 0 sinx( cosx + ) + ( cosx + ) 0 sinx + 0 or cosx + 0 sinx + cosx + 0 sinx or cosx sinx sinx sin x The general solution of this is 6 6 n n+ n + x n+ ( ) n+ ( ) x n and cosx cosx cos cos x The general solution of this is x n± i.e. x n± n + From () and (), we have n + and 6 n± are the required solutions (i)... (ii) 8. METHODS OF SOLVING TRIGONOMETRIC EQUATIONS 8. Factorization Trigonometric equations can be solved by use of factorization. Illustration : Solve: ( sinx cosx)( + cosx) sin x Sol: Use factorization method to solve this illustration. ( sinx cosx)( + cosx) sin x ( sinx cosx)( + cosx) ( cosx)( + cosx) 0 ; + cosx 0 or sin x 0 cos x or sin x cos x cos or sin x sin /6 sinx cosx + cosx sin x 0 x ( n + ),n I or x n+ ( ) n / 6,n I The solution of given equation is ( n + ),n I or n + cosx sinx 0 n+ / 6, n I

19 Mathematics Sum to Product Trigonometric equations can be solved by transforming a sum or difference of trigonometric ratios into their product. Illustration 5: If sin 5x + sin x + sin x 0 and 0 x /, then x is equal to. α+β α β Sol: By using sum to product formula i.e. sinα+ sinβ sin cos. sin 5x + sinx sin x sinxcosx + sinx 0 sinx 0,cosx / x n,x n ± ( /) sinx cosx + 0 Illustration 6: Solve cosx + sinx sinx 0 Sol: Same as above illustration, by using formula α+β α β sinα sinβ cos sin We can solve this illustration. cos x + sin x sin x 0 cosx + cosx.sin( x) 0 cosx cosx.sinx 0 cosx( sinx) 0 cosx 0 or sin x 0 x ( n + ),n I or sinx 6 x n+,n I 6 x ( n + ),n I or n Solution of given equation is ( n + ),n I or n 8. Product to Sum 6 n+,n I 6 Trigonometric equations can also be solved by transforming product into a sum or difference of trigonometric ratios. Illustration 7: The number of solutions of the equation sin5xcosx sin6xcosx, in the interval 0,, is: Sol: Simply by using product to sum method. The given equation can be written as ( sin8x + sinx) ( sin8x + sinx) sin x sin x 0 sin x cos x 0 Hence sin x 0 or cos x 0. That is, x n( n I), or x k+ ( k I). Therefore, since x 0,, the given equation is satisfied if x 0, 5,, or. 6 6 Hence, no. of solutions is 5.

20 7.0 Trigonometric Ratios, Identities and Equations 8. Parametric Methods General solution of trigonometric equation acos+ bsin c To solve the equation acos+ bsin c, put a r cos φ,b r sinφ such that Substituting these values in the equation, we have, r cosφcos+ r sinφsin c cos If c r ( φ ) cos( φ ) a c + b c > a + b, then the equation acos+ bsin c has no solution. b r a + b, φ tan a If c a + b, then put n a c + b cos α, φ ±α n±α+φ so that cos φ cosα Illustration 8: Solve: sinx + cosx Sol: Solve by using above mentioned parametric method. Given, cosx + sinx, dividing both sides by a + b cosx + sinx cos x cos 6 5 x n± x n± + x n+, n where n I 6 6 Note: Trigonometric equations of the form a sin x + b cos x c can also be solved by changing sin x and cos x into their corresponding tangent of half the angle. i.e ttan x/. The following example gives you insight. Illustration 9: Solve: cos x + sin x 5 x tan Sol: As we know, cosx x + tan will be get the result. and sinx x tan x + tan. Therefore by substituting these values and solving we cos x + sin x 5 (i) x x tan tan cosx & sinx Equation (i) becomes x x + tan + tan x x tan tan + 5 x x tan tan + + x Let tan t Equation (ii) becomes (ii) t t + 5 t + + t t t + 0 t 0 t / t tanx /

21 Mathematics 7. x x tan tan tanα, where tanα x n+α x n+ α where, α tan,n I 8.5 Functions of sin x and cos x Trigonometric equations of the form P (sin x ± cos x, sin x cos x) 0, where P (y, z) is a polynomial, can be solved by using the substitution sin x ± cos x C. Illustration 0: Solve: sin x + cos x + sin x. cos x Sol: Consider sin x + cos x t, and solve it by using parametric method. sin x + cos x + sin x. cos x Let sin x + cos x t sin x + cos x + sinx.cosx t Now, put sin x + cos x t and sin x. cos x t sinx.cosx t in (i), we get t + t t t + 0 t t sin x + cos x sinx + cosx (ii)... (i) Dividing both sides of equation (ii) by, we get: sinx + cosx. cos x cos If we take the positive sign, we get x n+, n I If we take the negative sign, we get x n, n I x n± 8.6 Using Boundaries of sin x and cos x Trigonometric equations can be solved by the use of boundness of the trigonometric ratios sinx and cos x. MASTERJEE CONCEPTS (i) The answer should not contain such values of angles which make any of the terms undefined or infinite. (ii) Never cancel terms containing unknown terms on the two sides, which are in product. It may cause loss of the general solution. Suppose the equation is sin x (tan x)/. Now, cancelling sinx on both the sides, we get only cos x, sin x 0 is not counted. (iii) Check that the denominator is not zero at any stage while solving equations. (iv) While solving a trigonometric equation, squaring the equation at any step must be avoided if possible. If squaring is necessary, check the solution for extraneous values. Suppose the equation is sin x sin x. We know that the only solution of this is sin x 0 but on squaring, we get (sin x) (sin x) which is always true. (v) Domain should not change, it if changes, necessary corrections must be made. Shivam Agarwal (JEE 009, AIR 7)

22 7. Trigonometric Ratios, Identities and Equations Illustration : Solve: sin x + cos x (JEE ADVANCED) Sol: By using boundary condition of sin x and cos x. Since sinx and cosx, we have, sinx + cosx Thus, the equality holds true if and only if sinx and cosx x n n + and x n± i.e. n n x + and x n± 6 Solution set is, n x x n + x x n± 6 Note: Here, unlike all other problems, the solution set consists of the intersection of two solution sets and not the union of the solution sets. x x Illustration : sinx cos sinx + + sin cosx ( cosx) 0. Find the general solution. (JEE ADVANCED) Sol: Open all brackets of given equation and then by using sum to product formula and method of finding general solution we will get the result. x x sinxcos sin x + cosx + sin cosx cos x 0 x sin x + + cosx 5x sin 5x n x,0,8 5x sin cosx + sin 5x and cos x x n + 5 ; cos x x m +..AP x ( m ) m I x + m 8, n I Illustration : Find the general solution of sin x + + 8sinxcos x (JEE ADVANCED) Sol: First square on both side and then using sum and difference formula we can solve this illustration. sin x + + 8sinxcos x sinx cosx + sin x cos x + + sinxcosx + 8sinxcos x sin x + cos x + sin6x + 8sinxcos x + sin 6x 8 sin x cos x + 8sinxcos x + sin 6x sin x cos x + sin 6x (sin 6x + sin x) sin x sin x ½ x sin x cos x + n x + n I 5 9. SIMULTANEOUS EQUATIONS Two equations are given and we have to find the value of variable which may satisfy both the given equations,

23 Mathematics 7. like cos cosα and sin sinα So, the common solution is n+α,n I Similarly, sin sinαand tan tanα So, the common solution is n+α, n I Illustration : The most general value of satisfying the equations cos and tan is: Sol: As above mentioned method we can find out the general value of. cos cos 9 7 n± ; n I Put n, tan tan n /, n I The common value which satisfies both these equation is 7 Hence, the general value is n+. 7 Put n, ; Put n, 7. Illustration 5: The most general value of satisfying equations sin and tan / are: Sol: Similar to above illustration. We shall first consider values of between 0 and sin sin or sin( / 6) 7 /6, /6 ; tan / tan ( / 6) tan ( + / 6) /6,7 /6 Thus, the value of between 0 and which satisfies both the equations is 7 /6. Hence, the general value of is n+ 7 / 6 where n I (a) Any formula that gives the value of PROBLEM SOLVING TACTICS A sin in terms of sin A shall also give the value of sin n n+ A. A (b) Any formula that gives the value of cos in terms of cos A shall also give the value of cos n ± A. (c) A Any formula that gives the value of tan in terms of tan A shall also give the value of tan n ± A.

24 7. Trigonometric Ratios, Identities and Equations (d) If α is the least positive value of which satisfies two given trigonometric equations, then the general value of will be n+α. For example, sin sinα and cos cosα, then, n +α,n I (i) sin( n+ ) ( ) n sin, n I (ii) cos( n+ ) ( ) n cos, n I (iii) n sin n sin, n I FORMULAE SHEET Tangent and cotangent Identities sin tan cos and cos cot sin Product Identities sin cosec, cos sec, tan cot Pythagorean Identities sin + cos, tan + sec, + cot csc Even/Odd Formulas Periodic Formulas (If n is an integer) Double and Triple Angle Formulas sin cot ( ) sin, cos( ) cos, tan tan ( ) cot, sec( ) sec, cosec, cosec ( + ) sin, cos( n ) cos +, sec( n+ ) sec, cosec( n ) ( ) sincos, sin sin sin sin n cot(n ) cot sin cos( ) cos sin tan tan( ), tan +, tan(n + ) tan, cos cos cos tan tan tan tan + cosec Complementary angles sin ± cos, cos ± sin, tan ± cot, cot tan, sec cosec, cosec sec Half Angle sin cos( ), cos cos( ) + cos( ), tan + cos( ) sin( α±β ) sinαcosβ± cosαsinβ, Sum and Difference cos( α±β ) cosαcosβ sinαsinβ, tanα± tanβ tan( α±β ), tanαtanβ

25 Mathematics 7.5, sinαsinβ cos( α β) cos( α+β) Product to Sum sinαcosβ sin sin α+β + α β, cosαcosβ cos cos α β + α+β, cosαsinβ sin sin α+β α β, α+β α β Sum to Product sinα+ sinβ sin cos, α+β α β sinα sinβ cos sin α+β α β cosα+ cosβ cos cos α+β α β cosα cosβ sin sin